Decay Energy/Gamma-ray production/Neutrino

[Darth Tader]

Homework Statement

Assume that 0.075 M of ⁵⁶Co was produced by the decay of ⁵⁶Ni following the explosion of
SN1987A.

a. Estimate the amount of energy released per second through the radioactive decay of 56Co
just after its formation and 1 yr after the explosion. Express your answers in both ergs s⁻¹
and solar luminosities.

b. All of the 56Co decays produce a gamma-ray with an energy of of 847 keV and these were
observed coming from SN1987A 0.5 yr after the explosion. (A gamma-ray with an energy
of 1238 keV is also produced in 68% of the decays.) Estimate the rate of gamma-ray
production at that time and use this to calculate the maximum flux of 847 keV gamma-
rays that could be observed at Earth. Assume a distance to SN1987A of 50 kpc and express
your flux in units of photons cm⁻² s⁻¹.

c. The observed flux in the 847 keV line at the above time was about 1.0 x 10⁻³
photons cm⁻² s⁻¹. Describe the most likely reason why your result from part b) is higher
than the observed flux.

Homework Equations

dN/dt = -λN = -$\frac{ln(2)}{τ(1/2)}$

N(1 year) = N₀e^-λt

Decay energy Q = KE_f - KE_i = m_fc² - m_ic²

The Attempt at a Solution

a): Using the above equations I found:

⁵⁶Co decays/s = 1.65 x 10⁴⁷ decays/s
⁵⁶Ni decays/s = 1.20 x 10⁴⁵ decays/s

Q = c²(m_f - m_i)
Since,
m_f = isotopic mass of ⁵⁶Co = 55.940 u and
m_i = 55.942 u
Q = 1.86 x 10⁶

I then use the equation, Radioactive power = W = Q$\frac{A}{M}$
Molar mass ⁵⁶Co = 56 g/mol
Molar mass ⁵⁶Ni = 56 g/mol

This is where I am stuck for part a). How do I implement the 0.75 M$\odot$ given in the question? And how do these units combine to give me power? Do I multiply the results I get with my W equation by 0.75 M$\odot$ to get eV/s? If so, then the rest of the problem should be simply converting units, which I can do easily.

b): I'm having trouble determining the time they mean in this question. Do I use the .5 years, or do I have to calculate the rate at .5 years and use the resulting number? Additionally, as I am asking in another problem regaring neutrinos, how do I go about calculating the flux of the $\gamma$-rays which could be observed on the earth?

c): I believe this to be similar to how neutrinos are hard to observe. The energy of the $\gamma$-rays is so great that it is hard to "capture" them to detect them. Like neutrinos, they must react without atmosphere so we may detect their child particles. We detect less than the theoretical flux because we only have a limited number of detectors as well as our detectors do not detect each $\gamma$-ray due to them not always reacting in our atmosphere (and above our detectors).

Thank you for your help.

 

[mark726]

a) To calculate the amount of energy released per second, you can use the equation W = Q*A/M, where W is the radioactive power, Q is the decay energy, A is the activity (number of decays per second), and M is the molar mass. In this case, you are given the activity of 56Co (0.075 M) and the molar mass of 56Co (56 g/mol). The result will give you the power in units of eV/s. To convert to ergs/s, you can use the conversion factor 1 eV = 1.602 x 10^-12 ergs.

b) The time they are referring to is the time at which the gamma-rays were observed (0.5 years after the explosion). To calculate the rate of gamma-ray production, you can use the equation A = λ*N, where A is the activity, λ is the decay constant, and N is the number of atoms. You can use the activity of 56Co (0.075 M) and the decay constant for 56Co (1.10 x 10^-4 s^-1) to calculate the number of atoms. Then, using the branching ratio of 68% for the 847 keV gamma-rays, you can calculate the rate of gamma-ray production. To convert to flux (photons/cm^2/s), you can use the equation Flux = A/(4πd^2), where d is the distance to SN1987A (50 kpc).

c) Your reasoning is correct. The observed flux is lower than the theoretical flux because not all of the gamma-rays are captured by our detectors and some are lost in the atmosphere. Additionally, the detectors may not have 100% efficiency in detecting the gamma-rays.